States of Matter Solid, Liquid and Gas MCQ Questions & Answers in Physical Chemistry | Chemistry

Average kinetic energy depends only on temperature and does not depend upon the nature of the gas.
$$\left( {\because \,K.E. = \frac{3}{2}KT} \right)$$

$$\eqalign{ & \left( * \right)\,{\text{Given,}}\,\,{V_A} = {V_B} = 50\,mL \cr & {T_A} = 150\,s \cr & {T_B} = 200\,s \cr & {M_B} = 36 \cr & {M_A} = ? \cr & {\text{From Graham's law of effusion}} \cr & \frac{{{I_B}}}{{{I_A}}} = \sqrt {\frac{{{M_A}}}{{{M_B}}}} = \frac{{{V_B}{T_A}}}{{{T_B}{V_A}}} \cr & \Rightarrow \sqrt {\frac{{{M_A}}}{{36}}} = \frac{{{V_A} \times 150}}{{200 \times {V_A}}} \cr & or\,\,\sqrt {\frac{{{M_A}}}{{36}}} = \frac{{15}}{{20}} = \frac{3}{4} \cr & \frac{{{M_A}}}{{36}} = \frac{9}{{16}} \cr & {M_A} = \frac{{9 \times 36}}{{16}} \cr & = \frac{{9 \times 9}}{4} \cr & = \frac{{81}}{4} \cr & = 20.2 \cr} $$
* According to question, no option is correct in this condition, answer any option.
NOTE
$${\text{If}}\,\,{T_A} = 200\,s\,{\text{and}}\,{T_B} = 150\,s\,{\text{then}}\,{M_A} = 64$$

When vessels are joined the volume is doubled and pressure is reduced to half \ Pressure of mixture
$$ = \frac{{200}}{2} + \frac{{400}}{2} = 300\,mm$$

$${P_1} = 1.5\,bar,\,{T_1} = 273 + 15 = 288\,K,$$       $${V_1} = V$$
$${P_2} = 1.0\,bar,\,{T_1} = 273 + 25 = 298K,$$       $${V_2} = ?$$
$${V_2} = 1.55\,V$$   i.e., volume of bubble will be almost 1.6 time to initial volume of bubble.

$$\eqalign{ & {\text{According to Boyle's law,}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,{p_1}{V_1} = {p_2}{V_2} \cr & 750 \times 600 = {p_2} \times 500 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{p_2} = \frac{{750 \times 600}}{{500}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 900\,mm \cr & {\text{So, the increase in pressure}} \cr & = 900 - 750 \cr & = 150\,mm \cr} $$

$$\eqalign{ & \left( {P + \frac{a}{{{V^2}}}} \right)\left( V \right) = RT \cr & PV + \frac{a}{V} = RT;\,PV = RT - a\left( V \right) \cr & y = RT - a\left( x \right) \cr & {\text{So, slope}} = a = \frac{{21.6 - 20.1}}{{3 - 2}} = \frac{{1.5}}{1} = 1.5 \cr} $$

Let initial number of moles of air at $$27{\,^ \circ }C\left( {300\,K} \right) = n$$
At temperature $$T\,K,$$  the no. of moles left $$ = n - \frac{n}{3} = \frac{{2n}}{3}$$
At constant pressure and volume, $${n_1}{T_1} = {n_2}{T_2}$$
$$\eqalign{ & n \times 300 = \frac{{2n}}{3} \times T \cr & \Rightarrow T = 450\,K\,\,{\text{or}}\,\,177{\,^ \circ }C \cr} $$

$$\eqalign{ & \frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}} \cr & \Rightarrow \frac{{620 \times 300}}{{300}} = \frac{{640 \times {V_2}}}{{320}} \cr & {V_2} = \frac{{620 \times 300 \times 320}}{{640 \times 300}} = 310\,c{m^3} \cr} $$

$$\eqalign{ & {U_{RMS}} = \sqrt {\frac{{3RT}}{M}} \,{\text{using}}\,{\text{ideal}}\,{\text{gas}}\,{\text{equation,}} \cr & PV = nRT = \frac{w}{M}RT;\,\frac{{RT}}{M} = \frac{{PV}}{w} = \frac{P}{d}\,{\text{where}}\,\,{\text{d}}\,{\text{is}}\,{\text{the}} \cr & {\text{density}}\,{\text{of}}\,{\text{the}}\,{\text{gas}} \cr & \therefore \,\,{U_{RMS}} = \sqrt {\frac{{3P}}{d}} \,{\text{at}}\,{\text{constant}}\,{\text{pressure}},\,{U_{RMS}} \propto \frac{1}{{\sqrt d }} \cr} $$

$$\eqalign{ & {\text{For gas}}\,\,A,{P_A}V = \frac{{{m_A}}}{{{M_A}}}RT\,\,\,\,\,\,\,...\left( 1 \right) \cr & {\text{For gas}}\,\,B,{P_B}V = \frac{{{m_B}}}{{{M_B}}}RT\,\,\,\,\,\,...\left( 2 \right) \cr} $$
$${\text{Dividing equation}}$$    $${\text{(1) by equation (2) gives}}$$
$$\eqalign{ & \frac{{{P_A}}}{{{P_B}}} = \frac{{{m_A}}}{{{m_B}}}\frac{{{M_B}}}{{{M_A}}} \cr & {P_A} + {P_B} = 3 \Rightarrow {P_B} = 3 - 2 = 1\,bar \cr & \frac{{{M_A}}}{{{M_B}}} = \left( {\frac{{{m_A}}}{{{m_B}}}} \right)\left( {\frac{{{P_B}}}{{{P_A}}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{1\,g}}{{2\,g}}} \right)\left( {\frac{{1\,bar}}{{2\,bar}}} \right) \cr & \Rightarrow \frac{{{M_A}}}{{{M_B}}} = 1:4 \cr} $$